# Mathematics: 4^x=8x proof dat x=2?

Posted: 01-May-2015 [20:06:41] by Igboanugo Okwudili

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4'x = 8x -----eq1
2'2x = (2³)x
x = 2'2x/2³
x = 2'(2x-3)

sub x in eq1
4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))
2'(4(2x-3)) = 2'(3+2x-3)
base cancels out
4(2x-3) = 2x
8x - 12 = 2x
6x =12
x = 2 = Final answer!
13-May-2015 22:48:44 ·
#Opeyemi; hmmm,pls which rule says dat u can make x a subj of d formular and x will still b a variable factor in wat u make it 2 be????if it a new stuff pls enligthen me...
20-May-2015 10:15:57 ·

4^x=8x
soln: 4^x=8x
4^x=x(2^2+2^2)
4^x=x(4)^2
4^x=x(2)^2×2
2^2x=2x^4
2x=4
dividing both sides by 2.
x= 2
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12-May-2015 00:04:26 ·
Igboanugo Okwudili; How.i can't understand wot u did
18-May-2015 08:36:13 ·
4^x=8x,
(1 3)^x=8x,
1 3x x(x-1)/2=8x,
1 3x (x^2-x)/2=8x,
(2 6x x^2-x)/2=8x,
2 6x x^2-x=16x,
x^2 6x-x-16x 2=0,
x^2-11x 2=0. Solve the equation quadratically,u will get 2.
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12-May-2015 18:26:44 ·
berber; +plus is at empty spaces
12-May-2015 18:39:08 ·
#Opeyemi; guy u no u can use addition sign in it na....(1+3)^x is not equal to (4)^x....so please i think u are wrong!
20-May-2015 10:18:27 ·
berber; You said that (1+3) is not equal to (4)?, you have go to primary school for better foundation.
30-Oct-2015 08:18:12 ·
berber; Mistake! There is 3^2 behind x(x-1)/2., it will now be 1+3x+9x(x-1)/2=8x,
which later gives 9x^2-19x+2=0 as the quadratic equation .
08-Nov-2015 14:23:24 ·
Mahmmudlawal; @ berber your proved is explicitly wrong as (1+3)^x is not equal to 4^x
You can't just do maths anyhow. You can verify by put x=3
09-Nov-2016 22:03:29 ·
berber; You need to go back to primary school for better foundation. Telling me that (1+3)=(4) is not correct
15-Nov-2016 22:01:18 ·
4'x = 8x -----eq1
2'2x = (2³)x
x = 2'2x/2³
x = 2'(2x-3)

sub x in eq1
4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))
2'(4(2x-3)) = 2'(3+2x-3)
base cancels out
4(2x-3) = 2x
8x - 12 = 2x
6x =12
x = 2 = Final answer!
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13-May-2015 22:48:44 ·
#Opeyemi; hmmm,pls which rule says dat u can make x a subj of d formular and x will still b a variable factor in wat u make it 2 be????if it a new stuff pls enligthen me...
20-May-2015 10:15:57 ·
4^x=8x
log2^2x=log8 + logx ( tak log to base 2 )
2xlog2 = 3log2 + logx
2x=3 + logx
2x = 3 + logx/log2 ( change log to base x)
2x = 3 +1/log2 ( log nw in base x)
2x log2/log2= 3 log2/log2 + 1/log2 (as log2/log2=1)
2x log2/log2=(3log2 + 1)/log2 (simplify rhs)
2x log2= 3logx +1
2xlog2 -3log2=1
log2(2x-3)=1 ( rem log is in base x)
2(2x-3)=x
4x-6=×
3x=6
x=2 (as required)
if any mistake , error or mathz error feel free to corect me. nobody is an highland of knwledge
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19-May-2015 01:51:00 ·
Igboanugo Okwudili; Bro u tried bt u take log to base 2 1st den to base x without using d formular to the change of base.xo i prove u wrong 4 dat
19-May-2015 17:02:54 ·
Rhylex; tanx but pls help me change logx(base2) to log in base x
20-May-2015 11:18:09 ·
noahonimisi; pls (2x-3)log2 is nt equal to log2(2x-3) ..therefor U re wrong
16-Dec-2016 00:58:44 ·
X=2 if not still contented with cross check by substitute x=2 in the equation.
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19-May-2015 21:12:55 ·
Igboanugo Okwudili; I said prove milionism
20-May-2015 00:36:25 ·
using logbase2.... logbase2 (4^x) =logbase2 (8x) 2xlogbase2 (2)=logbase2 (8)+logbase2 (x) 2x=3logbase2 (2)+logbase2 (x) 2x=3 + logbase2 (x) 2x-3=logbase2 (x) 2^(2x-3)=x^1 2=x and 2x-3=1 [2x-3=1;2x=4;x=2]
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20-May-2015 10:57:43 ·
#Opeyemi; pls if u dnt understand ma working ,u can view ma pic..
20-May-2015 11:05:22 ·
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07-Nov-2015 [22:05:54] ·
berber; Using binomial theorem.
4^x=8x,
(1+3)^x=8x,
1+3x+x(x-1)3^2/2=8x,
1+3x+(9x^2-9x)/2=8x,
(2+6x+9x^2-9x)/2=8x,
9x^2-3x+2=16x,
9x^2-3x-16x+2=0,
9x^2-19x+2=0,
(9x-1)(x-2)=0,
9x-1=0 or x-2=0,
9x=1 or x=2,
x=1/9 or x =2,
therefore x=2(it satisfy the equ.).
08-Nov-2015 14:15:37 ·
Hey guys, to solve that you'll have know a particular rule thats,
LogaN=logbN/logbA.
So, 4'x=8x
XLog(base 2)4=3log(base 2)2 + log(base 2)x. (you'll get that after simplifying)
then,
2x=3 +log(base10)x/log(base10)2
2x=3+x/2
4x=6+x
3x=6, x=2.
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22-May-2016 20:39:35 ·
A 19 year old boy named Ayanlaja Ramoni Adebola from Nigeria has just discovered is own mathematical formula for solving mathematical problem of a^x=bx. A question was asked by his friend to him which brought his research to find the solution to the mathematical problem. He was asked to find the value of x in the equation 4^x=8x.He had to make research on such questions hoping they exist. But he discovered that most people know that the answer was 2 but what is the solution(workings) to the problem? Some made use of graph method, some made use of Eisten Raph method but he noticed most people had to limit the value of 2 because they already knew the answer was 2.
He made his research on such question and concluded there must be workings leading to the solution of the problem.
By his handwork he proposed his own formula which he named "Debo's formula". Which states that for the equation a^x=bx, then x={log{b/loga)} /loga. (where x is approximated to a whole number).
Chek out the solution of 4^x=8x using Debo's formula.
4^x=8x
Solution
a=4 and b=8
Debo's formula ==>x=log{b/loga} divided by log a
X=log{8/log4} /log4
X=log{8/0.6021} /log4
X=log 13.2868/log4
X=1.1234/0.6021
X=1.86=2(Approximate to the nearest whole number)
Thus, using Debo's formula x=2.
Proof that x=2 from Debo's formula
4^x=8x
4^2=8*2
16=16
His friends argued that his formula can only be used to solve 4^x=8x...Another question of that type was created.
Solve 3^x=9x
Using Debo's formula
a=3 and b=9
x=log{b/loga} divided by loga
x=log{9/log3) divided by log3
x=log(9/0.4771) /log3
x=log18. 864/log3
x=1.2756/0.4771
x=2.67=3(approximate to nearest whole number)
X=3 using Debo's formula.
Proof ==>
3^x=9x
3^3=9*3
27=27
Using Debo's formula
Another problem was created
8^x=32x
Using Debo's formula
a=8 and b=32
Debo's formula ==>x=log{b/loga} divided by loga
x=log{32/log8} /log8
x=log {32/0.9031}/log 8
X=log 35.43/log8
X=1.5494/0.903
X=1.7=2(Approximate to the nearest whole number)
Using Debo's formula x=2
Proof that x=2
8^x=32x
8^2=32*3
64=64
The conditions in using Debo's formula are both the LHS and RHS numeric value must be both even or both odd. Also the numeric value at exponential part (a^x) must be lesser and be divisible by the numeric value of the side with identity x. Thus in a^x=bx, a
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09-Jul-2016 21:13:03 ·
4^x = 8x
Introduce natural logathrims to both sides
Log4^x = log8x
From loga + logb = logab
.: log4^x = log8 + logx
Express the numbers in their simplest form
Log2^2x = log2^3 + logx
apply the rule of power law
2xlog2 = 3log2 + logx
By Rearranging the terms
logx = 2xlog2 - 3log2
by factorizing the expression
Logx = log2(2x - 3)
by dividing through by log2
Logx/log2 = (2x - 3)
Note loga/logb = a/b
Therefore we have,
x/2 = (2x - 3)
Multiply through by 2
x = 4x - 6
By collecting their like term,
-3x = -6
Divide both sides by -3
X = 2 ..........................proved
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11-Jul-2016 17:20:15 ·
Thanks Guys bt i have not seen any meaningful proof yet
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07-Nov-2016 22:44:48 ·
4^x=8x
Divide through by x
Then, 4^x/x =8
4^x/x =16/2 since 16/2 =8
Equate either the numerator or denominator
4^x =16 =4^2
X=2
Proved.......
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09-Nov-2016 21:50:17 ·
This question seems to have no true solution...4^x=8x. And if there is I would like to know it.
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18-Nov-2016 15:59:30 ·
Igboanugo Okwudili; The answer is two. Bt to prove it is the problem
19-Nov-2016 08:41:17 ·
4^2=8(2)
4*4=8*2
16=16
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17-Dec-2016 08:21:11 ·